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3 years ago

A line passes through the two points: (1, 10) and (6,5).

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

5 0

Answer:

(a). - 1 ; (b). y = - x + 11

Step-by-step explanation:

(1, 10)

(6, 5)

(a). m = $\frac{y_{2} -y_{1} }{x_{2} -x_{1} }$ = $\frac{5-10}{6-1}$ = - 1

(b). y - 5 = - 1(x - 6)

y = - x + 11

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The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

8 0

2 years ago

PLEASE HELP ME PICTURE PROVIDED

egoroff_w [7]

Answer:

B and F

Step-by-step explanation:

B. ∠ j and ∠k

F. ∠i and ∠t

4 0

3 years ago

Read 2 more answers

jessica weighs x+34 pounds and Ronda weighs 12 pounds less. If Jessica gains 5 pounds and Ronda loses 2 pounds, what is the sum

Vladimir [108]

Answer:

2x+59.

Step-by-step explanation:

Let J represent Jessica's weight and R represent Ronda's weight.

Jessica weighs x+34 pounds. Thus:

$J=x+34$

Ronda weighs 12 pounds less than Jessica. In other words:

$R=J-12=(x+34)-12\\R=x+22$

The sum of their weights, therefore, is:

$J+R\\=(x+34)+(x+22)=2x+56$

Now, if Jessica gains 5 pounds and Ronda loses 2 pounds, the net gain of the total weight would be 3 pounds. Thus, we only need to add 3 to the original total to find the sum of their new weights:

$2x+56+3=2x+59$

The sum of the new [weights] is represented by 2x+59.

8 0

2 years ago

∠ABD is formed by a tangent and a secant intersecting outside of a circle. If minor arc AC = 78° and minor arc CD = 124°, what i

lana [24]

Minor arc AD is 360° -124° -78° = 158°. Then angle ABD is
(158° -78°)/2 = 40°

The appropriate selection is
C) 40°

4 0

2 years ago

Read 2 more answers

In the third week of July, a random sample of 40 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. A

suter [353]

Answer:

a. [6.6350,7.3950]

b. ME=0.5150

Step-by-step explanation:

a. Given that n=40, $\bar x=6.88$ and that: $z_{\alpha/2}=z_{0.05}=1.645$

The required 90% confidence interval can be calculated as:

$\bar x\pm(margin \ of \ error)\\\\\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\6.88\pm(1.645\times \frac{1.98}{\sqrt{40}})\\\\=[6.3650,7.3950]$

Hence, the 90% confidence interval for the population mean cash value of this crop is [6.6350,7.3950]

b. The margin of error at 90% confidence interval is calculated as:

$ME= z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\\\=(1.645\times \frac{1.98}{\sqrt{40}})\\\\=0.5150$

Hence, the margin of error is 0.5150

8 0

3 years ago